A bakery makes doughnuts and sprinkles half of the surface area with sugar. If k and l are the prices of dough and sugar. Minimize the total cost. Given that the doughnut is a torus with minor radius $r$ and major radius $R$ the surface area $A = 4 R r \pi^2$ and the volume $V= 2R\pi^2r^2$ with $R=ar^{-3/2}$. If $k$ is the price of dough and $l$ the price of sugar, minimise the total cost. (0<r<R)
For the total cost $K$ I found that $K=kV+\dfrac{lA}{2}=2k\pi^2Rr^2+2l\pi^2Rr$.
after substituting R and differentiating we get: $\dfrac{dK}{dr}=a\pi^2\dfrac{rk-l}{r^{\frac{3}{2}}}$.
After setting to zero and solving for $r$, I get $r=\dfrac{l}{k}$ for the minimum value of $K$ but I'm not certain if that is correct. It seems strange that the ideal radius reduces to a ratio of prices.
 A: Calculation is correct, but perhaps not strange to have ideal/optimal radius  r as ratio of costs.
It is just by chance of the exponent value being $-\frac{3}{2}$ that leads to the coincidence. Another exponent will result in another $r$ value for optimum condition.
To establish this I assumed a constraint condition linking $(r,R)$ more generally as
$$ R= a r^n $$
and proceeding essentially same way for minimum cost ( I used Euler Lagrange Equation for optimum ) and obtained ( for all coefficient $a$ values)  the optimum minor radius $r$ :
$$ r=- \frac{l}{k}\cdot \frac {n+1}{n+2} $$
$$ \text{ When n takes on values }\left(- \frac{3}{2},- \frac{5}{4} \right) \text { we get $r$ values}\left( \frac{l}{k}, \frac{l}{3k} \right); $$
Moreover the ratio of prices is not non-dimensional, but is linear $inch$ length dimension for the inner radius. The way you wrote total cost, sugar price is for spreading on the doughnut's area and dough price is for filling up the dough volume.
$$  \frac{l}{k}= \frac{$/in^2}{$/in^3}= [in]. $$
